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Resonance and SHM

I have summarised the question, with all the key Points. However if this is not clear let me know please.

Question:

Whilst driving at 70kmh-1, the car goes over undulating stretch of road. Driver notices the car bounced up dramatically. Calculate how far apart the undulations in the road are. If effect was due to resonance of the car suspension.

thats pretty much the whole question.
Additional information: Spring constant is 25000Nm-1

would i have to use f=kx and solve for x?
Reply 1
Original post by joyoustele
I have summarised the question, with all the key Points. However if this is not clear let me know please.

Question:

Whilst driving at 70kmh-1, the car goes over undulating stretch of road. Driver notices the car bounced up dramatically. Calculate how far apart the undulations in the road are. If effect was due to resonance of the car suspension.

thats pretty much the whole question.
Additional information: Spring constant is 25000Nm-1

would i have to use f=kx and solve for x?


Resonance occurs if the natural frequency matches applied force, have u been given the mass of the car?
Reply 2
Original post by Shaanv
Resonance occurs if the natural frequency matches applied force, have u been given the mass of the car?


Yes. IN a previous question.
Reply 3
Original post by joyoustele
Yes. IN a previous question.


Use ur formula relating time period, mass and spring constant. This is the time between bumps.

V=st, with ur time period will give u the required distance.

In future can u post a picture of the full question its easier for us to help u👊🏾
Reply 4
Original post by Shaanv
Use ur formula relating time period, mass and spring constant. This is the time between bumps.

V=st, with ur time period will give u the required distance.

In future can u post a picture of the full question its easier for us to help u👊🏾


Okay, Thank you Shaanv
Reply 5
Original post by joyoustele
Okay, Thank you Shaanv


No worries mate.
Reply 6
Original post by Shaanv
Use ur formula relating time period, mass and spring constant. This is the time between bumps.

V=st, with ur time period will give u the required distance.

In future can u post a picture of the full question its easier for us to help u👊🏾


I got 25.9 m, im not sure if thats correct.

I found the time period to be 1.33s

Using S=ut

S=displacement
U=Speed
t=time

S=(70*1000)/3600 * 1.33

26M seems quite large
Reply 7
Original post by joyoustele
I got 25.9 m, im not sure if thats correct.

I found the time period to be 1.33s

Using S=ut

S=displacement
U=Speed
t=time

S=(70*1000)/3600 * 1.33

26M seems quite large


From what i see i have no problems with ur working out, the only possible source of error could be ur time period, as i dont know the mass i cannot double check this.

The car moves at about 19.4 m/s so in 1.33 seconds u would expect about 26 metres.

But that distances sounds right for the figures i have access to.

👊🏾
Reply 8
Original post by Shaanv
From what i see i have no problems with ur working out, the only possible source of error could be ur time period, as i dont know the mass i cannot double check this.

The car moves at about 19.4 m/s so in 1.33 seconds u would expect about 26 metres.

But that distances sounds right for the figures i have access to.

👊🏾


Love you man! Thanks so much.:smartass::tee:
Reply 9
Original post by joyoustele
Love you man! Thanks so much.:smartass::tee:


Everyone apart from me has got such cool emojis on this site😢. I need to fix up my game

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